From Förberedande kurs i matematik 1

After moving the terms over to the left-hand side, so that

\displaystyle \sqrt{2}\sin x\cos x-\cos x=0

we see that we can take out a common factor \displaystyle \text{cos }x,

\displaystyle \cos x\left( \sqrt{2}\sin x-1 \right)=0

and that the equation is only satisfied if at least one of the factors, \displaystyle \text{cos }x or \displaystyle \sqrt{2}\text{sin }x-\text{1} is zero. Thus, there are two cases:

\displaystyle \text{cos }x=0: This basic equation has solutions \displaystyle x={\pi }/{2}\; and \displaystyle x=3{\pi }/{2}\; in the unit circle, and from this we see that the general solution is

\displaystyle x=\frac{\pi }{2}+2n\pi and \displaystyle x=\frac{3\pi }{2}+2n\pi

where \displaystyle n\text{ } is an arbitrary integer. Because the angles \displaystyle {\pi }/{2}\; and \displaystyle 3{\pi }/{2}\; differ by \displaystyle \pi , the solutions can be summarized as

\displaystyle x=\frac{\pi }{2}+n\pi ( \displaystyle n an arbitrary integer).

√ \displaystyle \sqrt{2}\text{sin }x-\text{1}=0

if we rearrange the equation, we obtain the basic equation as

\displaystyle \text{sin }x\text{ }={1}/{\sqrt{2}}\;, which has the solutions \displaystyle x={\pi }/{4}\; and \displaystyle x=3{\pi }/{4}\; in the unit circle and hence the general solution

\displaystyle x=\frac{\pi }{4}+2n\pi and \displaystyle x=\frac{3\pi }{4}+2n\pi

where \displaystyle n\text{ } can arbitrary integer.

All in all, the original equation has the solutions

\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{\pi }{4}+2n\pi \\ x=\frac{\pi }{2}+n\pi \\ x=\frac{3\pi }{4}+2n\pi \\ \end{array} \right. ( \displaystyle n\text{ } an arbitrary integer).